ap chemistry 2014-2015 ch 4 types of chemical reactions and solution stoichiometry ch 3/4 quiz t...
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AP Chemistry 2014-2015
CH 4 TYPES OF CHEMICAL
REACTIONS AND SOLUTION
STOICHIOMETRYCh 3/4 Quiz T 9/16Ch 3/4 Exam Th 9/18
Water has a high specific heat, high heat of vaporization, and high adhesive/cohesive forces
The two O-H bonds in water are polar covalent (oxygen is more electronegative—partial negative charge on oxygen, partial positive charge on each hydrogen) Bond angle =~105° (the two unshared
electron pairs on oxygen are “space hogs”, squishing the bond angle from a tetrahedral 109.5 despite four areas of electron density)
4.1 WATER, THE COMMON SOLVENT
Partially positive hydrogens are attracted to negative ions, partially negative oxygen is attracted to positive ions hydration
4.1 CONTINUED
For ionic solids: when the hydration attraction are greater than the crystal lattice attractions, the compound is soluble
Water can also dissolve nonionic substances (ex. alcohols, sugars) if they exhibit polarity
Fats (generally all nonpolar substances) do not dissolve in water
“Like dissolves like”
4.1 CONTINUED
A solution is homogeneous mixture where a solute is dissolved in a solvent; if the solvent is water the solution is considered “aqueous”
Properties of aqueous solutionsElectrolytes are solutions that conduct an electric current Strong electrolytes completely dissociate (ex. strong acids,
strong bases, soluble salts) Common strong acids: HCl, HBr, HI, HNO3, H2SO4, HClO4
Common strong bases: oxides and hydroxides of group 1A and 2A metals (2A metal salts tend to be less soluble than 1A metal salts)
Weak electrolytes do not completely dissociate (only about 1% dissociation)—examples include acetic acid and ammonia
4.2 THE NATURE OF AQUEOUS SOLUTIONS: STRONG AND WEAK
ELECTROLYTES
Nonelectrolytes are solutions in which the solute dissolves but does not make ions. They cannot conduct electricity. Ex. pure water, sugar, alcohol, antifreeze, starch
Arrhenius—the extent to which a solution can conduct an electric current depends on the number of ions present
AQUEOUS SOLUTIONS CONTINUED
Molarity = moles of solute/liters of solution (ex. 0.75 M NaCl means that for every 1 L of solution there are 0.75 moles of NaCl present)
4.3 THE COMPOSITION OF SOLUTIONS
Calculate the molarity of a solution prepared by dissolving 11.5 g of solid NaOH in enough water to make 1.50 L of solution. (answer = 0.192 M)
EXERCISE 4.1 MOLARITY I
11.5 g / 40.00 g/mol = 0.288 mol NaOH
M = moles/volume = 0.283 moles / 1.50 L = 0.192 M
Calculate the concentration of the cobalt (II) ions and the nitrate ions in a 0.50 M solution of cobalt (II) nitrate. (answer = 0.50 M Co2+
and 1.0 M NO3-)
EXERCISE 4.2 MOLARITY II—IONS
0.50 M Co(NO3)2
[Co2+] = [Co(NO3)2] = 0.50 M
[NO3-] = 2[Co(NO3)2] = 1.00 M
Calculate the number of moles of Cl - ions in 1.75 L of 0.001 M ZnCl2. (answer = .0035 moles Cl -)
EXERCISE 4.3 MOLES AND MOLARITY
1.75 L 1 L 0.0001 moles
1 mole ZnCl2
2 moles Cl- = 0.0035 moles Cl-
Typical blood serum is about 0.14 M NaCl. What volume of blood contains 1.0 mg NaCl? (answer = 0.12 mL)
EXERCISE 4.4 VOLUME AND MOLARITY
1.0 mg NaCl
58.44 g1000 mg
1 g NaCl 1 mole NaCl 1 L0.14 moles
= 1.2 x 10-4 L
A standard solution is a solution whose concentration is accurately known. To prepare a solution of known concentration, weigh out the solid as accurately as possible and place it in a volumetric flask. Add enough just enough distilled water to dissolve the solid and then fi ll to the mark on the flask. Mix.
STANDARD SOLUTIONS
To analyze the alcohol content of a certain wine, a chemist needs 1.00 L of an aqueous 0.200 M potassium dichromate solution. How many grams of solid K2Cr2O7 are needed to make this solution?
EXERCISE 4.5 STANDARD SOLUTIONS
1.00 L1 L0.200 moles
1 mole K2Cr2O7
294.20 g = 58.8 g K2Cr2O7
M1V1 = M2V2
Generally, you measure out the quantity of stock (concentration) solution that you calculate from the formula above, and place it in a volumetric flask (of the correct volume for your dilution). Next you fill the volumetric flask to the mark on the neck with distilled water until the meniscus is reached. Mix.
DILUTION
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M sulfuric acid solution? (answer = 9.4 mL)
EXERCISE 4.6 DILUTION
(16 M)(x L) = (0.10 M)(1.5 L)
x = 0.0094 L or 9.4 mL
Review terms: chemical reaction, balanced chemical equation, (s), (l), (g), (aq)
Not all reactions fall neatly into one category
4.4 TYPES OF CHEMICAL REACTIONS
The formation of a precipitate is the driving force for some chemical reactions. A precipitate is an insoluble solid that is formed when two aqueous solutions are mixed. We can separate the precipitate from solution by filtration in what is called gravimetric analysis.
4.5 PRECIPITATION REACTIONS
Most alkali metal salts and ammonium salts are soluble
Chloride, bromide, and iodide are soluble—except for Ag+, Hg2
2+, and Pb2+
Nitrates, chlorates, perchlorates, and acetates are soluble
Sulfates are soluble—except for Ca2+, Sr2+, Ba2+, Ag+, Hg2
2+, and Pb2+
Carbonates, phosphates, chromates, dichromates, sulfi des, hydroxides, and oxides are insoluble—but the fi rst rule (most alkali metal salts and ammonium salts are soluble) takes priority
It can be assumed that ionic compounds that dissolve in water are strong electrolytes.
YOU HAVE TO KNOW THE SOLUBILITY RULES
Predict what will happen when the following pairs of solutions are mixed. Potassium nitrate and barium chloride
Sodium sulfate and lead (II) nitrate
Potassium hydroxide and iron (II) nitrate
EXERCISE 4.7 SOLUBILITY RULES
2 KNO3(aq) + BaCl2(aq) 2 KCl(aq) + Ba(NO3)2(aq)No reactionNa2SO4(aq) + Pb(NO3)2(aq) PbSO4(s) + 2 NaNO3(aq)Formation of a precipitate, lead (II) sulfate2 KOH(aq) + Fe(NO3)2 (aq) 2 KNO3(aq) + Fe(OH)2(s)No reaction
Complete balanced equation—gives the overall reaction stoichiometry but does not specify which reactants/products ionize
Complete ionic equation—same as above but represents all strong electrolytes as ions
Net ionic equation—includes only the solution components undergoing a change; spectator ions are not includedThere is always a conservation of charge in net ionic equations
4.6 DESCRIBING REACTIONS IN SOLUTION
Aqueous potassium chloride is added to aqueous silver nitrate to form a silver chloride precipitate plus aqueous potassium nitrate.
Write the complete balanced equation:
Write the complete ionic equation:
Write the net ionic equation:
EXERCISE 4.8 THREE TYPES OF EQUATIONS
KCl(aq) + AgNO3(aq) AgCl(s) + KNO3(aq)
K+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) AgCl(s) + K+
(aq) + NO3-(aq)
Cl-(aq) + Ag+(aq) AgCl(s)
Precipitation reactions form an insoluble salt; this salt is dried and its mass is obtained
We can use stoichiometry to determine the expected/needed quantities of each compound
4.7 STOICHIOMETRY OF PRECIPITATION REACTIONS
Calculate the mass of solid sodium chloride that must be added to 1.50 L of a 0.100 M silver nitrate solution to precipitate all of the silver ions in the form of AgCl. (answer = 8.77 g NaCl)
EXERCISE 4.9 PRECIPITATION REACTION STOICHIOMETRY I
1.50 L AgNO3
1 L AgNO3
0.100 moles AgNO3
1 mole AgNO3
1 mole NaCl
1 mole NaCl
58.44 g NaCl
= 8.77 g NaCl
NaCl(aq) + AgNO3(aq) AgCl(s) + NaNO3(aq)
When aqueous solutions of sodium sulfate and lead (II) nitrate are mixed, lead (II) sulfate precipitates. Calculate the mass of lead (II) sulfate formed when 1.25 L of 0.0500 M lead (II) nitrate and 2.00 L of 0.0250 M sodium sulfate are mixed. (answer = 15.2 g lead (II) sulfate)
EXERCISE 4.10 PRECIPITATION REACTION STOICHIOMETRY II
Na2SO4(aq) + Pb(NO3)2(aq) PbSO4(s) + 2 NaNO3(aq)1.25 L Pb(NO3)2
1 L
0.0500 moles Pb(NO3)2 1 mole
Pb(NO3)2
1 mole PbSO4
1 mole PbSO4
303.3 g PbSO4
= 19.0 g PbSO4
2.00 L Na2SO4
1 L
0.0250 moles Na2SO4 1 mole
Na2SO4
1 mole PbSO4
1 mole PbSO4
303.3 g PbSO4
= 15.2 g PbSO4
Acids are compounds that produce anions and hydronium ions (H3O+) when they react with water (you can also just say hydrogen ions, H+, though hydronium is technically more correct)—Arrhenius defi nition
Bases are compounds that produce cations and hydroxide ions when they react with water—Arrhenius defi nition NH3 is therefore not an Arrhenius acid, but it IS a Bronsted-
Lowry acid because it is a proton acceptorIn a neutralization reaction, an acid and a base in
equimolar (really equinormal) quantities react to produce water and a salt. The salt may or may not be soluble.
4.8 ACID-BASE REACTIONS
What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL of 0.350 M NaOH? (answer = 0.0875 L = 87.5 mL HCl)
EXERCISE 4.11 NEUTRALIZATION
HCl(aq) + NaOH(aq) H2O(l) + NaCl(aq)In other words, HCl is monoprotic and NaOH is monobasic, so they react in a 1:1 ratio. M1V1 = M2V2 applies (not always true for polyprotic acids/polybasic bases).(0.100 M HCl)(V1 mL) = (0.350 M NaOH)(25.0 mL) V1 = 87.5 mL
Volumetric analysis is a technique for determining the amount of a certain substance by doing a titration
Review terms: titrant, analyte, burette, indicator
ACID-BASE TITRATIONS
The equivalence point occurs when the number of moles of hydroxide ions is equal to the number of moles of hydrogen (hydronium) ions—this term is also used in redox titrations
The end point is where you see a color change when using an indicator
Standardization is a procedure for establishing the exact concentration of a reagent—ex. you may prepare a solution to be 0.50 M but it might be “off”; standardization helps you to determine the true concentration
A student carries out an experiment to standardize a sodium hydroxide solution. To do this the student weighs out a 1.3009 g sample of potassium hydrogen phthalate (KHC8H4O4 abbreviated KHP, molar mass 204.22 g/mol, has one acidic hydrogen). The student dissolves the KHP in distilled water, adds phenolphthalein as an indicator, and titrates the resulting solution with the sodium hydroxide solution to the phenolphthalein endpoint. The difference between the final and initial burette readings is 41.20 mL. Calculate the concentration of the sodium hydroxide solution. (answer = 0.1546 M NaOH)
EXERCISE 4.12 STANDARDIZATION
EXERCISE 4.12 CONTINUED
KHP: mass = 1.3009 g molar mass = 204.22 g/mole NaOH: volume = 41.20 mL molarity = ?
KHP is monoprotic, so KHP and NaOH react in a 1:1 ratio.Moles KHP = moles NaOH
Moles KHP = 1.3009 g / 204.22 g/mole = 0.00637 moles KHP
Moles NaOH = volume * molarity = 0.04120 L * V = .4120V
0.00637 = 0.04120V V = 0.1546 M
Terms to knowOIL RIG (oxidation is loss, reduction is gain, of electrons)
Oxidation is the loss of electrons, which increases charge (more positive)
Reduction is the gain of electrons, which decreases charge (more negative)
Oxidation number is the assigned charge on an atom
Oxidizing agents get reduced, reducing agents get oxidized—reduction and oxidation must be coupled
4.9 OXIDATION-REDUCTION REACTIONS
Always zero for elementsThe oxidation state of a monatomic ions is the same as its
chargeFluorine is always -1, oxygen is almost always -2
(exceptions—peroxides where it is -1, or OF 2 where it is +2)
Hydrogen is almost always +1; metal hydrides are an exception, where it is -1 (in these situations, hydrogen is placed at the end of a chemical formula like LiH)
The sum of the oxidation states must be zero for a neutral compound; for polyatomic ions, the sum of the oxidation states must equal the charge on the polyatomic ion
It’s odd, but there can be non-integer oxidation states. Ex. Fe3O4 where the oxygens total -8, so iron’s charge must be +8/3 Fe+8/3
RULES FOR ASSIGNING OXIDATION STATES
Assign oxidation states to each atom in the following compounds/ions.
CO2 SF6 NO3-
C = S = N =
O= F = O =
EXERCISE 4.13 OXIDATION STATES
+4 +5
-2
+6
-1-2
For the following reaction, identify the atoms that are oxidized and reduced.
2 Al(s) + 3 I2(s) 2 AlI3(s)
Now go back and identify the oxidizing and reducing agents .
EXERCISE 4.14 OXIDATION AND REDUCTION I
Aluminum: 0 +3, so aluminum was oxidizedIodine: 0 -1, so iodine was reduced
Aluminum: reducing agentIodine: oxidizing agent
The following reactions are associated with metallurgy. For each reaction, identify the atoms that are oxidized and reduced. Then identify the oxidizing and reducing agents. 2 PbS(s) +3 O2(g) 2 PbO(s) + 2 SO2(g)
PbO(s) + CO(g) Pb(s) + CO2(g)
EXERCISE 4.15 OXIDATION AND REDUCTION II
Ox: S (-2 +4) Red: O (0 -2)
Ox: C (+2 +4) Red: Pb(+2 0)
Ox agent: oxygen Red agent: sulfur
Ox agent: lead Red agent: carbon
Divide the equation into oxidation and reduction half reactions (OIL RIG)
Balance all elements besides hydrogen and oxygen
Balance oxygen by adding water to the appropriate side of the equation
Balance hydrogen by adding hydrogen ions to the appropriate side of the equation
4.10 BALANCING OXIDATION-REDUCTION EQUATIONS
Balance the charge by adding electronsMultiply the half reactions to make the
electrons equal for oxidation/reduction reactions
Cancel terms when you recombine the two half reactions
These rules are for acidic solutions; if this takes place in a basic solution, you have one more step. Neutralize any hydrogen ions by adding the same number of hydroxide ions to each side; check your water and cancel terms as necessary
Check
4.10 CONTINUED
Balance the following equation using the half-reaction method (acidic):MnO4
-(aq) + I -(aq) Mn2+(aq) + I2 (aq)
Reduction: Oxidation: Combined:
SAMPLE PROBLEM 1
MnO4- Mn2+ + 2 H2O4 H+ +
Mn: +7 +2; gained 5 electrons
5 e- +
2 I- I2
I: -1 0; each iodine atom lost 1 electron (2 total)
+ 2 e-
Need to multiply Mn equation by 2 and I equation by 5 to get 10 electrons transferred; the number of electrons lost in one equation must equal the number of electrons gained in the other equation
Manganese: 8 H+ + 10 e - + 2 MnO4- 2 Mn2+ + 4 H2O
Iodine: 10 I - 5 I2 + 10e -
SAMPLE PROBLEM 1 CONTINUED
Now, we add and cancel. In this equation, only the electrons will cancel.
8 H+ + 2 MnO4- + 10 I - 2 Mn2+ + 4 H2O + 5 I2
We would reduce if we could, but this equation is finished.
Balance the following equation using the half-reaction method (basic):Ag(s) + CN -(aq) + O2(g) Ag(CN)2
-(aq) + H2O
Reduction: Oxidation: Combined:
SAMPLE PROBLEM 2
O: 0 -2; gained 2 electrons on two atoms
Ag: 0 +1; lost 1 electron
4 H+ + O2 + 4 e- 2 H2O
Ag + 2 CN- Ag(CN)2- + e-
4 H+ + O2 + 2 Ag + 4 CN- 2 Ag(CN)2- + e-
+ 2 H2OThis equation is only correct in acidic solution; let’s make it basic now
4 H+ + O2 + 2 Ag + 4 CN - 2 Ag(CN)2- + e - + 2 H2O
To make this basic, we will add enough hydroxide ions to each side to neutralize the four hydrogen ions present on the products side. This will give us4 OH - + 4 H+ + O2 + 2 Ag + 4 CN - 2 Ag(CN)2
- + e - + 2 H2O + 4 OH -
Which simplifies to2 H2O + O2 + 2 Ag + 4 CN - 2 Ag(CN)2
- + e - + 4 OH -
Because we made four molecules of water on the reactants’ side and we had two molecules of water on the right, we have to cancel two molecules of water from each side.
Potassium dichromate reacts with ethanol (C2H5OH) as follows: H+(aq) + Cr2O7
2-(aq) + C2H5OH(l) Cr3+(aq) + CO2(g) + H2O(l)
Balance this equation using the half-reaction method.
EXERCISE 4.15 BALANCING HALF REACTIONS
Cr: +6 +3 so 3 electrons are gained; 2 atoms present14 H+ + 6e- + Cr2O7
2- 2 Cr3+ + 7 H2O
C: -2 +4 so 6 electrons are lost; 2 atoms present3 H2O + C2H5OH 2 CO2 + 12 e- + 12 H+
We have 6 electrons gained in one reaction and 12 electrons lost in the other. We can fix this by multiplying so that 12 electrons are transferred (x2 chromium equation)
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